Tavas and Karafs CodeForces - 536A

Karafs is some kind of vegetable in shape of an 1 × h rectangle. Tavaspolis people love Karafs and they use Karafs in almost any kind of food. Tavas, himself, is crazy about Karafs.Each Karafs has a positive integer height. Tavas has an infinite 1-based sequence of Karafses. The height of the i-th Karafs is si = A + (i - 1) × B.For a given m, let's define an m-bite operation as decreasing the height of at most m distinct not eaten Karafses by 1. Karafs is consideredas eaten when its height becomes zero.Now SaDDas asks you n queries. In each query he gives you numbers l, t and m and you should find thelargest number r such that l ≤ r and sequence sl, sl + 1, ..., sr canbe eaten by performing m-bite no more than t times or print -1 if there is no such number r.InputThe first line of input contains three integers A, B and n (1 ≤ A, B ≤ 106, 1 ≤ n ≤ 105).Next n lines contain information about queries. i-th line contains integers l, t, m(1 ≤ l, t, m ≤ 106)for i-th query.OutputFor each query, print its answer in a single line.ExampleInput

2 1 4
1 5 3
3 3 10
7 10 2
6 4 8

Output

4
-1
8
-1

Input

1 5 2
1 5 10
2 7 4

Output

1
2

题意：

给一个以a为首相，公差为b的等差数列，然后有n组询问；

比如2 1 4：等差数列s为：2,3,4,5,6,7,8,9,10，...

然后询问 ：l,t,m，就是从等差数列的第l项开始，操作t次，每次可以操作至多m个数，令这m个数减一；

问是否存在一个数字r(r>=l)，使得[l,r]之间的所有项都减为0;

例如：

1 5 3    操作五次，最多每次减三个数

开始：2,3,4,5,6,7.....

第一次：2,2,3,4,6,7.....

第二次：2,1,2,3,6,7.....

第三次：2,0,1,2,6,7.....

第四次：1,0,0,1,6,7.....

第五次：0,0,0,0,5,7.....

所以r最多等于r；

思路：

这是一道水题：二分就可以过了；

我们只需要二分满足两个条件，即：

等差数列s[r] < t && s[l,r]的和小于t*m即可

具体代码如下：

#include <bits/stdc++.h>#define LL long longint main() {LL ans;LL a, b, n;LL l, t, m;while(~scanf("%lld%lld%lld",&a, &b, &n)) {while(n--) {scanf("%lld%lld%lld",&l, &t, &m);if(t < (a+(l-1)*b)) printf("-1\n");else {LL L = l , R = (t-a)/b + 1;while(L <= R) {LL M = L+(R-L)/2;LL dn = M-l+1;//前n项和公式 n*a1+n*(n-1)/2;if(dn*(a+b*(l-1)) + b*dn*(dn-1)/2 <= t*m) {L = M+1;}else R = M-1;}printf("%lld\n",L-1);}}}return 0;}